Solve:16⋅42−33⋅1 \sqrt{16}\cdot4^2-3^3\cdot\sqrt{1} 16⋅42−33⋅1
Begin by evaluating the square roots: 16=4 \sqrt{16} = 4 16=4 and 1=1 \sqrt{1} = 1 1=1.
Substitute these back into the expression:
4⋅42−33⋅1 4\cdot4^2-3^3\cdot1 4⋅42−33⋅1
Calculate each term:
42=16 4^2 = 16 42=16, so 4⋅16=64 4\cdot16 = 64 4⋅16=64
33=27 3^3 = 27 33=27, so 27⋅1=27 27\cdot1 = 27 27⋅1=27
Subtract the second result from the first:
64−27=37 64 - 27 = 37 64−27=37
60